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insens350 [35]
2 years ago
7

A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted

on selected samples, but all components are also subjected to Vickers hardness testing. (A) What HV (in Kgf/mm2 ) is the minimum acceptable value? (B) If hardness is measured using Rockwell C hardness test, what is the minimum HRC value that is acceptable?
Engineering
1 answer:
Pavlova-9 [17]2 years ago
8 0

Answer:

a) The minimum acceptable value is 387.5 HV using Vickers hardness test.

b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.

Explanation:

To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.

According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):

\mbox{Tensile strength}=HV*3.2\\\\HV  = \frac{\mbox{Tensile strength}}{3.2} =\frac{1240}{3.2}=387.5

According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:

\mbox{Tensile strength}=HRC*31.5\\\\HRC  = \frac{\mbox{Tensile strength}}{31.5} =\frac{1240}{31.5}=39.4

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Darina [25.2K]

Answer:

0.1047N

Explanation:

To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s

N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N

in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047

3 0
3 years ago
Assume the availability of an existing class, ICalculator, that models an integer arithmetic calculator and contains: an instanc
shtirl [24]

We connect with computers through coding, often known as computer programming.

<h3>How to code?</h3>
  • We connect with computers through coding, often understood as computer programming.
  • Coding exists similar to writing a set of instructions because it instructs a machine what to do.
  • You can instruct computers what to do or how to behave much more quickly by learning to write code.

class ICalculator {

int currentValue;

int add(int value) {

this.currentValue = currentValue + value;

return currentValue;

}

int sub(int value) {

this.currentValue = currentValue - value;

return currentValue;

}

int mul(int value) {

this.currentValue = currentValue * value;

return currentValue;

}

int div(int value) {

this.currentValue = currentValue / value;

return currentValue;

}

}

public class ICalculator2 extends ICalculator {

int negate() {

if (currentValue != 0)

this.currentValue = -currentValue;

return currentValue;

}

public static void main(String[] args) {

ICalculator2 ic = new ICalculator2();

ic.currentValue=5;

System.out.println(ic.add(2));

System.out.println(ic.sub(5));

System.out.println(ic.mul(3));

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System.out.println(ic.negate());

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To learn more about code, refer to

brainly.com/question/22654163

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3 0
1 year ago
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\

\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

For cylinder BC:

Let Length of BC = 18 in

c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\

\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998}  =1.3266*10^{-6}T_{BC}

2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

8 0
3 years ago
Mohr's circle represents: A Orientation dependence of normal and shear stresses at a point in mechanical members B The stress di
blsea [12.9K]

Answer:

The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members

Explanation:

Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

\sigma _{x'x'}=\frac{\sigma _{xx}+\sigma _{yy}}{2}+\frac{\sigma _{xx}-\sigma _{yy}}{2}cos(2\theta )+\tau _{xy}sin(2\theta )

And \tau _{x'x'}=-\frac{\sigma _{xx}-\sigma _{yy}}{2}sin(2\theta )+\tau _{xy}cos(2\theta )

Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.

The Mohr circle is graphically displayed in the attached figure.

4 0
2 years ago
Its been days wsgggggg
Nataly [62]

Answer: ok

Explanation:

4 0
2 years ago
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